osmotic pressure is proportional to the molarity, C of the solution at a given temperature T. Thus: Π = C R T 
Here Π is the osmotic pressure and R is the gas constant. 
Π = (n₂ /V) R T 
Here V is volume of a solution in litres containing n2 moles of solute. If w2 grams of solute, of molar mass, M2 is present in the solution, then n₂ = w₂ / M₂ and we can write,

 in a dilute aqueous solution molarity is equal to molality.
c = m when p = 1 and solution is dilute.

$\mathrm{\pi } = {\mathrm{n}}_{\mathrm{B}}\frac{\mathrm{RT}}{\mathrm{V}} = \mathrm{cRT}=\mathrm{mRT}$


The osmotic pressure will increase with an increase in molality of the solution at a given temperature.

Answer:

osmotic pressure is proportional to the molarity, C of the solution at a given temperature T. thus

Π = C R T 
Here Π is the osmotic pressure and R is the
gas constant.
Π = (n2 /V) R T 
Here V is volume of a solution in litres containing n2 moles of solute.
If w2 grams of solute, of molar mass, M2 is present in the solution, then
n2 = w2 / M2 and we can write

$\mathrm{\pi V} = \frac{{\mathrm{w}}_2\mathrm{RT}}{{\mathrm{M}}_2}$

Thus, knowing the quantities w2, T, Π and V we can calculate the molar mass of the solute.

Answer:

(i) Liquid in liquid: Ethanol dissolved in water

(II) Gas in gas : Mixture of oxygen and nitrogen gases

van’t Hoff factor modifies the equations for colligative
properties as follows:
Relative lowering of vapour pressure of solvent,

$\frac{{\mathrm{p}}_1^0 - {\mathrm{p}}_1}{{\mathrm{p}}_1^0} = i\frac{n_2}{n_1}$

Elevation of Boiling point, ΔTb = i Kb m
Depression of Freezing point, ΔTf = i Kf m
Osmotic pressure of solution, Π = i n2 R T / V
 thus it shows that van't hoff factor depend on solute such that it is a colligative properties.

Van’t Hoff introduced a factor ‘i’ called Van’t Hoff’s factor, to express the extent of association or dissociation of solutes in solution. It is ratio of the normal and observed molecular masses of the solute, i.e


$\mathrm{i} =\frac{ \mathrm{Normal} \mathrm{molecular} \mathrm{mass}}{\mathrm{observed} \mathrm{molecular} \mathrm{mass}}$

In case of association, observed molecular mass being more than the normal, the factor i has a value less than 1. But in case of dissociation, the Van’t Hoff’s factor is more than 1 because the observed molecular mass has a lesser value than the normal molecular mass. In case there is no dissociation the value of ‘i’ becomes equal to one.